特殊情况：

• □ -> □：满足限制の二元限制个数 $v^2$

(1, 1), (1, 2), ..., (1, v),

(2, 1), (2, 2), ..., (2, v),

......

(v, 1), (v, 2), ..., (v, v)

• □ -> □ -> ...（m 条二元限制）-> □：满足限制の二元限制个数 $(v^2)^m$

• a -> b：不满足限制の二元限制个数 $(v-1)$

(a, 1), (a, 2), ..., (a, b-1), (a, b+1), ..., (a, v)

满足限制の二元限制个数 $v^2-(v-1) = v^2-v+1$

• a -> □ -> b：不满足限制の二元限制个数 $v*(v-1)$

(a, □) -> (□, 1), (□, 2), ..., (□, b-1), (□, b+1), ..., (□, v) $1 \le □ \le v$

满足限制の二元限制个数 $(v^2)^2-(v^2-v) = (v^2)^2-v^2+v$

一般情况：

• a -> □ -> □ -> ...（m 条二元限制）-> b：不满足限制の二元限制个数 $v^{m-1}*(v-1)$

满足限制の二元限制个数 $(v^2)^m-v^{m-1}*(v-1)$

P.S. 本题目略卡常。。。

#include<cstdio>
#include<map>
using namespace std;

int const mod=1e9+7;

long long pow(long long b, int p){
	long long ans=1;
	while(p){
		if(p%2) (ans *= b) %= mod;
		(b *= b) %= mod;
		
		p /= 2;
	}
	
	return ans;
}

map<int, int> num;

int fuction(){
	num.clear();

	int n, m;
	scanf("%d%d", &n, &m);
	
	int limit;
	scanf("%d", &limit);
	
	bool flag=0;
	while(m--){
		int ix, val;
		scanf("%d%d", &ix, &val);
		
		map<int, int>::iterator it=num.find(ix);
		
		if(it!=num.end() && it->second!=val) flag = 1;
		else num.insert(make_pair(ix, val));
	}
	
	if(flag) return 0;
	
	map<int, int>::iterator it=num.begin();
	long long ans=pow(pow(limit, 2), it->first-1);
	
	while(true){
		int pre=it->first;
        
        if(++it == num.end()) break;

        int now=it->first;
		(ans *= pow(pow(limit, 2), now-pre)-pow(limit, now-pre-1)*(limit-1)%mod+mod) %= mod;
	}

	(ans *= pow(pow(limit, 2), n-(--it)->first)) %= mod;
	
	return ans;
}

int main(){
    freopen("assign.in", "r", stdin);
    freopen("assign.out", "w", stdout);

	int T;
	scanf("%d", &T);
	
	while(T--){
		int ans=fuction();

		printf("%d\n", ans);
	}
	
    fclose(stdin);
    fclose(stdout);

	return 0;
}